Bending of Cyclist in Curves

Consider a cyclist negotiating a level (unbanked) circular road of radius $r$ at speed $v$. Treating the cyclist and cycle together as a single system of mass $m$ with center of gravity $C$, this system travels in a circle of radius $r$ about some center $O$; choose the line $OC$ as one axis and the vertical line through $O$ as another, to set up the geometry of the problem.

Working in the rotating frame. Because the system as a whole is rotating (going around the curve), the most natural way to analyse it is to work in a frame that co-rotates with the cyclist, in which the cyclist appears to be at rest. Since this rotating frame is non-inertial, Newton's laws only apply in it once an additional pseudo (centrifugal) force, of magnitude $\dfrac{mv^2}{r}$, is included, acting outward through the system's center of gravity. Four forces act on the system in this frame: (i) the gravitational force $mg$, acting downward through $C$; (ii) the normal force $N$, from the road, at the point of contact; (iii) the frictional force $f$, from the road, at the point of contact; and (iv) the centrifugal pseudo-force $\dfrac{mv^2}{r}$, acting outward through $C$. Since the system is in equilibrium in this rotating frame, the net force and the net torque must both be zero, exactly as for ordinary static equilibrium (§5.3.1).

Deriving the leaning angle. Take torques about the point of contact with the road, $A$. The gravitational force's torque, $mg\,(AB)$, causes a clockwise turn (taken as negative); the centrifugal force's torque, $\dfrac{mv^2}{r}(BC)$, causes an anticlockwise turn (taken as positive). Setting the net torque to zero:

From the geometry of the right triangle $ABC$ (with $\theta$ the angle the cyclist leans from the vertical), $AB=AC\sin\theta$ and $BC=AC\cos\theta$. Substituting:

So the required bending angle from the vertical is …